BUFFER SOLUTIONS Buffer solutions solution which can

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BUFFER SOLUTIONS

Buffer solutions solution which can resist the addition of a strong acid or a strong base or water. Its’ p. H changes very slightly. + • + 1 drop of base [H ] in 1000 times + • + 1 drop of acid [H ] in 5000 times -4 -7 (from 10 tо 5 х10 ) -7 In buffer solution from 1. 00 х10 -7 to 1. 01 х10

Classification Acidic Amfoteric Weak acid and its’ salt Aminoacids, proteins Acetate СН 3 СООН СН 3 СОО– (H 3 N+)m – Prot – (COOˉ)n НСО 3– Н 2 РО 4– НРO 42– Weak base and its’ salt Ammonia Hydrocarbonate Н 2 СО 3 Phosphate Basic НА Acid Н + + А- base Donor Аcceptor NH 4+ NH 3

Mechanism of buffer action Acetate buffer СН 3 СОО- + Na+ СН 3 СООNa СН 3 СООН СН 3 СОО- + Н+ + 1 mole Na. OH СН 3 СООН + ОН+1 mole HCL СН 3 СОО- + Н+ Н+ ОН 1 mole СН 3 СОО- + Н 2 О (weak electrolite ) СН 3 СООН 1 mole (weak electrolite)

р. Н formulas are derived from Kdis.

HOW TO PREPARE BUFFER 1. Mixing the components: -for acidic buffer p. H = p. Ka + lg Ns·Vs/Na·Va -for basic buffer p. H = 14 – p. Kв – lg Ns·Vs/Nb·Vb

2. Partial neutralization - For acidic buffer nacid = nbase = nsalt СН 3 СООН + Na. OH = CH 3 COONa + H 2 O (exsess) p. H = p. Ka + lg Nb·Vb / (Na·Va –Nb·Vb) - For basic buffer NH 4 OH + HCL = NH 4 Cl + H 2 O (exsess) p. H = 14 – p. Kв – lg Na·Va / (Nb·Vb - Na· Va)

Buffer capacity Ba = nacid / |∆р Н|. Vbuf. sol Вb = nbase/ |∆р Н| n – mole equivalents of a strong acid or a strong base Vbuf. sol - volume of a buffer solution ∆р. Н – p. H change as a result of acid or base addition

Buffer capacity depends on : 1. Components amount 2. nsalt/nacid or nsalt/nbase Вmax - for acidic buffer n at salt = nacid р. Н = р. Ка - for basic buffer at nsalt = nbase р. Н = 14 -р. Кb p. H = p. Ka + lg nsalt/nacid p. H = 14 - p. Kb - lg nsalt/nbase

Mechanism of buffer action Acetate buffer СН 3 СООNa СН 3 СООН СН 3 СОО- + Na+ СН 3 СОО- + Н+ Н+ ОН-

Buffer capacity nsalt > nacid Вa > Вb nsalt < nacid Вa < Вb nsalt = nacid Вa = Вb =Вmax p. H = p. Ka + lg nsalt/nacid

• Choose the buffer with maximum capacity and р. Н = 7. 36 : 1) acetic р. К = 4. 75; 2) phosphate р. К = 7. 21; 3) hydrocarbonate р. К = 6. 37.

Buffer systems of a body 1. Mineral Hydrocarbonate Phosphate Н 2 СО 3 НСО 3– Н 2 РО 4– НРO 42– 2. Protein and aminoacidic.

Hydrocarbonate buffer (K) Na. HCO 3/H 2 CO 3 H 2 O atmosphere СO 2 (gas) СO 2 (solution) lungs H 2 СO 3 H+ + HСO 3 - Blood plasma р. Н = p. Ka (H СO ) + lg C(Na. HCO )/C(H 2 CO 3) = 2 3 3 = 6, 1 + lg. C(HCO 3 -) – lg p(CO 2) p - CO 2 pressure in lungs Buffer capacity Вa = 40 ммole/L Вb = 1 -2 ммоle/L

р. Н of blood plasma 7. 4 = 6. 1 + lg [НСО 3–]: [СО 2] = 20: 1 –]/ [СО 2] Вa > Вb Н 2 СО 3 – 13 моle/ day Other acids – from 0. 03 to 0. 08 моle/ day

1. A buffer consists of 0, 5 moles of equivalent NH 3 and 0, 5 moles of equivalent NH 4 Cl. Which buffer component must be added to change p. H to 9? Kb(NH 3)=1, 8*10 -5 2. What is the p. H of buffer made of 60 ml of 0, 10 M NH 3 with 40 ml of 0, 10 M NH 4 Cl. Kb=1, 8*10 -5. 3. What volume of 0, 6 M CH 3 COONa must be added to 600 ml of 0, 2 M CH 3 COOH to produce a buffer with p. H=4, 75? Ka(CH 3 COOH)=1, 75*105.

4. What volume of 0, 01 M Na. OH should be added to 100 ml of 0, 5 M CH 3 COOH solution to produce a buffer with p. H 4, 75? p. Ka(CH 3 COOH)=4, 75 5. A buffer was prepared of 500 ml Na. Н 2 РО 4 and 500 ml Na 2 НРO 4 . After addition of 1 ml 0. 1 N HCl the change of buffer p. H = 0. 03. Calculate buffer capacity Ba. 6. Choose a buffer with Вa > Вb: a). 100 ml 0. 2 M Na. HCO 3 + 100 ml 0. 4 M H 2 CO 3 b). 100 ml 0. 4 M Na. HCO 3 + 100 ml 0. 2 M H 2 CO 3 c). 100 ml 0. 2 M Na. HCO 3 + 100 ml 0. 2 M H 2 CO 3

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